Deeper than primes

[jsfisher]

Poor jsfisher, he can't understand that ("completely covered by points") AND ("No two points are adjacent") is Option 1 of http://www.internationalskeptics.com/forums/showpost.php?p=6585117&postcount=12494.

No, that is your delusion, not mine.

Perhaps you were hung up by that word, additional? True enough, there is no additional space between any two distinct points. It is the same space that was always there, and that space is completely covered by points. Those points aren't additional, either, at least not in the sense of needing to be added. They have always been there.

The collection of points that covered a line is complete because it includes all the points that cover the line. Nothing was omitted.

By about fourth grade, this concept becomes understandable.

 

[doronshadmi]

Nothing was omitted.

jsfisher, you can't get that the size of 1-dimensional space is strictly greater than the size of collections of 0-dimensional spaces, even if they are uncountable, because you can't think beyond the concept of Collection.

The collection of points that covered a line is complete because it includes all the points that cover the line. Nothing was omitted.

So what? It is still a collection, and being a collection of 0-dimensional spaces is being strictly smaller that 1-dimensional space, which is a notion that you can't comprehend.

 

[doronshadmi]

1 dimensional space (a line) is by definition a collection of points (you are calling those 0 dimensional).

Only in your fantasy. Like jsfisher, you can't think beyond the concept of Collection.

 

[doronshadmi]

Those points aren't additional, either, at least not in the sense of needing to be added. They have always been there.

There is no process of any kind here jsfisher.

My heart with you when I see how your notion is closed under the concept of Collection, such that you can't get anything beyond it, but your notion's problems do not change the abstract notion of 1-dimensional space that is strictly greater than any collection of 0-dimesional spaces along it.

Furthermore 1-dimensional space exists even if there are no 0-dimesinal spaces along it and it has the magnitude of the continuum, but this notion is really beyond your head.

 

[HatRack]

Welcome HatRack,

Do you assert that a 1-dimensional space is completely covered by a collection of 0-dimensional spaces?

Please answer only by yes or no.

Doron, I will respond to your question when you respond to my original post. 1/2 + 1/4 + 1/8 + ... is equal to the limit at infinity of 1 - 1/2^n, according to the definition of the sum of an infinite series as found in any calculus text. From the standard definition of limits and all of the standard theorems which follow from them, which all in turn follow from the axioms of ZFC set theory given the appropriate set of definitions, we have 1/2 + 1/4 + 1/8 + ... = 1.

If you assert anything different, then you are calling one of the ZFC axioms or one of the traditional definitions found in analysis into question. I asked you to clearly and precisely state which axiom or definition you are changing to derive the proposition 1/2 + 1/4 + 1/8 + ... < 1.

 

[doronshadmi]

EDIT:

there is no additional space between any two distinct points. It is the same space that was always there, and that space is completely covered by points.

If 1-dimensional space is completely covered by 0-dimensional spaces, then 1-dimensional space does not exist.

In that case we are left with collection of distinct 0-dimensional spaces.

No two points are adjacent.

Again you are using top to bottom (macro) reasoning, which according to it no two points are adjacent.

But this is not the only possible reasoning, because by bottom to top (micro) reasoning, we are still deal with collection of distinct 0-dimesional spaces, and it this micro level a collection is complete only of there is exactly 0 distance between distinct 0-dimesional spaces, which is a contradiction, because 0-dimensional spaces can't be both distinct AND have 0 distances between them.

Since this is a contradiction, then given uncountable scale levels, it is always true that no two points are adjacent, such that there is always uncovered space between them, because being distinct is an essential property of any collection of 0-dimensional spaces.

This uncovered space is 1-dimensional space, which its magnitude is greater than the magnitude of the distinct 0-dimensional spaces along it.

 

[doronshadmi]

Doron, I will respond to your question when you respond to my original post. 1/2 + 1/4 + 1/8 + ... is equal to the limit at infinity of 1 - 1/2^n, according to the definition of the sum of an infinite series as found in any calculus text. From the standard definition of limits and all of the standard theorems which follow from them, which all in turn follow from the axioms of ZFC set theory given the appropriate set of definitions, we have 1/2 + 1/4 + 1/8 + ... = 1.

If you assert anything different, then you are calling one of the ZFC axioms or one of the traditional definitions found in analysis into question. I asked you to clearly and precisely state which axiom or definition you are changing to derive the proposition 1/2 + 1/4 + 1/8 + ... < 1.

HatRack,

ZFC is based on the concept Set, where a set is based on the concept of collection of distinct objects.

My question is essential to the concept of collection of distinct objects.

Please answer by yes or no to this essential question:

Do you claim that a collection of distinct 0-dimenssional spaces completely cover 1-dimensional space?

 

[HatRack]

HatRack,

ZFC is based on the concept Set, where a set is based on the concept of collection of distinct objects.

My question is essential to the concept of collection of distinct objects.

Doron, your line of reasoning is useless if it leads you to the proposition 1/2 + 1/4 + 1/8 + ... < 1. This is because it is ultimately derivable from the ZFC axioms, which give us the basic properties of sets, that 1/2 + 1/4 + 1/8 + ... = 1. By the trichotomy law, which is also derivable from ZFC, a number cannot simultaneously equal 1 and be less than 1. This means that, in the ZFC system alone, any proof you give that 1/2 + 1/4 + 1/8 + ... < 1 is flawed and needs no further examination.

Thus, the only way to reconcile your line of reasoning with standard mathematics is to change one of the ZFC axioms or change a definition so that the proposition 1/2 + 1/4 + 1/8 + ... = 1 is no longer derivable.

My question is simple and direct, and you still haven't answered it. Which ZFC axiom or definition are you modifying/abolishing to make way for your line of reasoning?

 

[doronshadmi]

It is the same space that was always there, and that space is completely covered by points.

Utter nonsense.

If that space (1-dimensional space) is completely covered by 0-dimensioanl spaces, then that space does not exist, so there is no "the same space that was always there".

No two points are adjacent

You are right, no two distinct 0-dimensional spaces are adjacent, and this abstraction is invariant upon any given infinitely many scale levels, which is exactly the 1-dimesional space, which its magnitude is strictly greater than any given collection of distinct 0-dimensioanal spaces along it.

 

[jsfisher]

jsfisher, you can't get that the size of 1-dimensional space is strictly greater than the size of collections of 0-dimensional spaces, even if they are uncountable, because you can't think beyond the concept of Collection.

As always, the limitation is yours, not mine.

EDIT:If 1-dimensional space is completely covered by 0-dimensional spaces, then 1-dimensional space does not exist.

What a bizarre comment. Baseless, untrue, and bizarre.

In that case we are left with collection of distinct 0-dimensional spaces.

Come on, you can say: points.

...it is always true that no two points are adjacent

Yes, I have said that before.

...such that there is always uncovered space between them

Maybe that's true in bizarro Doronetics world; in the land of real Mathematics, not so.

...because being distinct is an essential property of any collection of 0-dimensional spaces.

Great! You have thrown in yet another unrelated issue. Yes, all the points are distinct.

This uncovered space is 1-dimensional space, which its magnitude is greater than the magnitude of the distinct 0-dimensional spaces along it.

Go ahead. Show us any place along a line that is not covered by points. Any place at all. You keep claiming they are there. Heck, you keep claiming they are everywhere, yet you can't show us one. Why is that?

 

[jsfisher]

Utter nonsense.

Is it? Then why can't you show us any contradiction from within Mathematics? Instead, you resort to misrepresentations, gibberish, undefined terms, and leaps of reasoning with no basis in logic.

Come on, Doron. Stop heckling. Show us something real. How about showing us any place along a line where there isn't a point? Surely such a place must exist, because you told us so. Show it to us.

 

[epix]

The notation 1/2 + 1/4 + 1/8 + 1/16 + ... is, loosely speaking, alternative notation for the limit as n approaches infinity of the finite sum of the series 1/2 + 1/4 + ... + 1/2^n. This finite sum is of course 1 - 1/2^n. Hence, 1/2 + 1/4 + 1/8 + ... = lim n->inf (1 - 1/2^n) = 1. Doronshadmi's assertion that 1/2 + 1/4 + 1/8 + ... < 1 is complete nonsense for that reason. For any natural number n, it is true that 1/2 + 1/4 + 1/8 + ... + 1/2^n < 1. But, having a "+ ..." at the end indicates that the limit is being taken.

No. The notation 1/2 + 1/4 + 1/8 + 1/16 + ... is a descriptive and non-algebraic expression that merely establishes an infinite series. Since Doron is not familiar with the established symbolic language, he would use the description to indicate inequality 1/2 + 1/4 + 1/8 + 1/16 + ... < 1. This symbolism is normally avoided, coz it lacks necessary elements to prove or disprove the assertion. When this convergent series is converted into its algebraic form including the inequality

1 - 1/2ⁿ < 1 where n → ∞

it can be shown in three easy steps that the inequality holds.

1 - 1/2ⁿ < 1
-1/2ⁿ < 0
-1 < 0

However, the solution of the inequality is also good for any sum S ≥ 1, which allows the sum be greater than 1. That's why the limit needs to be computed. During the process

[lim n → ∞] 1 - 1/2ⁿ
the whole term 1/2ⁿ is held sufficiently close to zero to be removed under the rule that derives the limits. Obviously, the term never equals zero as n goes to infinity, and therefore the series cannot reach its limit.

No. The partial sums of the series are always less than 1 for any given natural number n. The infinite sum is defined as the limit of the sequence of partial sums, and the value of that limit is 1 in this particular case.

The conversion of the series into function f(x) = 1 - 1/2^x where x → ∞ leaves the issue of partial sums irrevelant, coz f(x) is a function and not a series comprising addends. But the "partiality" can be inherited and resurface in an assertion that cannot be easily defined. Let

f(x) = 1 - 1/2^x where x → ∞

and

g(x) = 1 - 1/2^x where x ≥ 1

In that case, f(x) ≠ g(x) as it is often asserted.

The concept of "partiality" must be properly translated into the function that substitutes the series and that's the point where it squeaks.

 

[epix]

Exactly!

But since Traditional Math asserts that a 1-dim space is completely covered by a collection of distinct 0-dim spaces, then (according to Traditional Math) two points along a 1-dim space are both distinct AND there is no gap between them.

In that case Traditional Math has to provide the formal proof, which is resulted by the ability of two 0-dim spaces along a 1-dim space, to be both distinct AND gap-less.

Let's see how Traditional Math formally proves it

There are two distinct points in "TRADITIONAL MATH" and that's R and N.

Isn't it enough?

No? Okay . . .

When N opens its door, these guys get out: 1, 2, 3, 4, ...
They are distinct and there is no gap between them.

 

[HatRack]

No. The notation 1/2 + 1/4 + 1/8 + 1/16 + ... is a descriptive and non-algebraic expression that merely establishes an infinite series.

I'm looking at a calculus textbook right now that agrees with me. It has this written almost exactly:

[latex]a + ar + ar^{2} + ... + ar^{n-1} + ... = \displaystyle\sum\limits_{n=1}^{\infty}ar^{n-1}=\lim_{n \to \infty}\displaystyle\sum\limits_{k=1}^{n}ar^{k-1}=\frac{a}{1-r}[/latex]

Substitute in 1 for a and 1/2 for r to get:

[latex]1 + 1/2 + 1/4 + ... + (1/2)^{n-1} + ... = \displaystyle\sum\limits_{n=1}^{\infty}(1/2)^{n-1}=\lim_{n \to \infty}\displaystyle\sum\limits_{k=1}^{n}(1/2)^{k-1}=\frac{1}{1-1/2}=2[/latex]

Subtract 1 from both sides and you have:

[latex]1/2 + 1/4 + 1/8 + ... = 1[/latex]

An analysis text I have from 1980 also uses the + ... to indicate the limit of the sequence of partial sums. It's not as precise as using the Sigma notation or using a limit symbol, but it's clear that it indicates an infinite summation, which is defined as the limit of the sequence of partial sums, which is 1 in this case. Hence, 1/2 + 1/4 + 1/8 + 1/16 + ... < 1 is nonsense.

If Doron defines 1/2 + 1/4 + 1/8 + 1/16 + ... in a nonstandard way, then he needs to specify that, which is what I've been asking him.

 

[doronshadmi]

Yes, all the points are distinct.

You are right jsfisher.

Being distinct point along 1-dimensional space is possible only if given any arbitrary pair of points, they are not adjacent.

You see jsfisher? No need to show any missing point, all we need is to understand how a given point is distinct along a given line, and it is distinct only if the distance to another point is > 0.

This is a simple abstract fact that can't be avoided, if all the points are distinct, exactly as you and me claim.

I understand my claim.

You do not undertand your claim.

 

[The Man]

You are right jsfisher.

Being distinct point along 1-dimensional space is possible only if given any arbitrary pair of points, they are not adjacent.

You see jsfisher? No need to show any missing point, all we need is to understand how a given point is distinct along a given line, and it is distinct only if the distance to another point is > 0.

This is a simple abstract fact that can't be avoided, if all the points are distinct, exactly as you and me claim.

I understand my claim.

You do not undertand your claim.

No Dorn you actually and simply don’t understand either claim. Yes the points are “distinct only if the distance to another point is > 0”, but guess what, (and this has been explained to you multiple times before in multiple ways including this one) any distance >0 can be divided by 2 giving at least one other distinct point between those two. In fact the whole infinite series can be mapped between those two points or any two points. Once again that is precisely what makes it a continuous space as opposed to a discrete space that you are claiming.

 

[jsfisher]

You are right jsfisher.

Tell my wife that.

Being distinct point along 1-dimensional space is possible only if given any arbitrary pair of points, they are not adjacent.

Very not true. Consider the pair of points on the real number line, 2 and 2. (No, you didn't specify the points needed to be distinct.)

You see jsfisher? No need to show any missing point, all we need is to understand how a given point is distinct along a given line, and it is distinct only if the distance to another point is > 0.

This is a simple abstract fact that can't be avoided, if all the points are distinct, exactly as you and me claim.

No two distinct points along a line are adjacent. Is that what you are trying to say? If so, then we all agree.

I understand my claim.

The evidence you have provided makes this doubtful. You dwell on the trivial then jump to the irrational without the constraint of logic. Heck, you haven't even figured out, yet, the difference between "two distinct points" versus "two adjacent points."

You do not undertand your claim.

Yep.

 

[jsfisher]

Yes, all the points are distinct.

You are right jsfisher.

By the way, the quote of mine was stripped of its context, and even in context, its true meaning was probably lost. At any rate, given A as a point on line L and B as a point on line L, there is no implicit requirement that A and B be two different points.

I believe I have been consistent, always adding the qualifier, distinct, where appropriate.

Given two arbitrary points (along a line or in a plane, or in 3-space, or ...):
(1) The points are not adjacent.
(2) If distinct, then there are uncountably many points along the line segment with the two points as open end-points.

Standby for a classic Doron misinterpretation in 5...4...3...

 

[The Man]

I believe I have been consistent, always adding the qualifier, distinct, where appropriate.

Quite true jsfisher, though I accidentally left that qualifier out in the third sentence of my last post.

 

[laca]

Only in your fantasy. Like jsfisher, you can't think beyond the concept of Collection.

So you got nothing? I understand.